Question

If the shortest distance between the lines $\frac{x-1}{\alpha}=\frac{y+1}{-1}=\frac{z}{1},$ (α ≠ -1) and x + y + z + 1 = 0 = 2x – y + z + 3 is $\frac{1}{\sqrt{3}}$, then a value α is:
(a) $-\frac{16}{19}$
(b) $-\frac{19}{16}$
(c) $\frac{32}{19}$
(d) $\frac{19}{32}$

Answer 1

Hello mate ☺

Option B is correct one

Hope it helps u...❤

Answer 2

If the shortest distance between the lines $\frac{x-1}{\alpha}=\frac{y+1}{-1}=\frac{z}{1},$ (α ≠ -1) and x + y + z + 1 = 0 = 2x – y + z + 3 is $\frac{1}{\sqrt{3}}$, then a value α is:

(a) $-\frac{16}{19}$

(b) $-\frac{19}{16}$✔

(c) $\frac{32}{19}$

(d) $\frac{19}{32}$